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Fluid kinemtics by basnayake mis
- 1. 1 Application of Bernoulli’s Equation: Flow measurement Measuring devices: The pitot tube and the pitot-static tube The venturi meter Orifices Bernoulli’s Equation 𝑝 + 𝑑𝑝 𝐴 + 𝑑𝐴 𝑝𝐴 𝑑𝑠 𝜃 𝜌𝑔𝐴 𝑑𝑠 Force momentum relation applied to steady, 1D, inviscid fluid flow 𝑑𝑧 Consider a small stream tube having length ds For an inviscid flow, shear stresses are absent 𝐹𝑠𝑦𝑠 = 𝜌𝑄 𝑉2 − 𝑉1 − 𝑝 + 𝑑𝑝 𝐴 + 𝑑𝐴 + 𝑝𝐴 − 𝜌𝑔𝐴 𝑑𝑠 𝑆𝑖𝑛𝜃 = 𝜌𝑄 𝑉2 − 𝑉1 −𝑑𝑝 𝐴 − 𝜌𝑔𝐴 𝑑𝑧 = 𝜌𝑄 𝑑𝑉 −𝑑𝑝 𝐴 − 𝜌𝑔𝐴 𝑑𝑧 = 𝜌𝐴 𝑉 𝑑𝑉 𝑑𝑝 + 𝜌𝑔 𝑑𝑧 + 𝜌 𝑉 𝑑𝑉 = 0Euler’s equation For a constant density fluid 𝑑 𝑝 + 𝜌𝑔𝑧 + 𝜌𝑉2 /2 = 0 𝑑 𝑝/𝜌𝑔 + 𝑧 + 𝑉2 /2𝑔 = 0 𝑝 + 𝜌𝑔𝑧 + 𝜌𝑉2 /2 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑝/𝜌𝑔 + 𝑧 + 𝑉2 /2𝑔 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Total pressure is constant along a streamline Total head is constant along a streamline or 1D Frictionless flow in a converging pipe 𝜃 Datum z=0 Total head Piezometric head line 𝑃1 𝜌𝑔 𝑉1 2 2𝑔 𝑍1 𝑍2 𝑃2 𝜌𝑔 𝑉2 2 2𝑔 𝐴1 𝐴2 𝑃 𝜌𝑔 + 𝑧 + 𝑉2 2𝑔 = 𝐻 Consider a stream of uniform velocity flows into a blunt body Stream line pattern Apply Bernoulli equation to central streamline (1) and (2) 𝑃1 𝜌𝑔 + 𝑉1 2 2𝑔 + 𝑍1 = 𝑃2 𝜌𝑔 + 𝑉2 2 2𝑔 + 𝑍2 𝑉2 = 0; 𝑍1 = 𝑍2 𝑃1 𝜌𝑔 + 𝑉1 2 2𝑔 = 𝑃2 𝜌𝑔 𝑃2 = 𝑃1 + 1 2 𝜌𝑉1 2 Stagnation Pressure Static Pressure Dynamic Pressure Pitot tube
- 2. 2 A piezometer and a pitot tube Two piezometers, one as normal and one as a Pitot tube within the pipe can be used in an arrangement shown below to measure velocity of flow 𝑃2 = 𝑃1 + 1 2 𝜌𝑉1 2 𝜌𝑔ℎ2 = 𝜌𝑔ℎ1 + 1 2 𝜌𝑉1 2 𝑉1 = 2𝑔 ℎ2 − ℎ1 Pitot tube 1 2 Pitot tube Pitot-static tube Close-up of a Pitot-static tube Pitot-static tube A pitot tube is connected to a manometer. The holes on the side of the tube connect to one side of a manometer and register the static head, (h1), while the central hole is connected to the other side of the manometer to register, as before, the stagnation head (h2). 1 2 𝜌𝑉1 2 = 𝜌𝑔ℎ − 𝜌 𝑚𝑎𝑛 𝑔ℎ 𝑉1 = 2𝑔ℎ 𝜌 𝜌 − 𝜌 𝑚 The Pitot/Pitot-static tubes give velocities at points in the flow Pitot-static tube (1) (1) (2) A B h x Connected point (1) Connected point (2) Consider the pressures on the level of the centre line of the Pitot tube 𝑃1 + 𝜌𝑔𝑥 = 𝑃2 + 𝜌𝑔(𝑥 − ℎ) + 𝜌 𝑚𝑎𝑛 𝑔ℎ 𝑃2 = 𝑃1 + 𝜌𝑔ℎ − 𝜌 𝑚𝑎𝑛 𝑔ℎ 𝑃2 = 𝑃1 + 1 2 𝜌𝑉1 2 Applying Bernoulli equation
- 3. 3 Bernoulli equation: 𝑃1 𝜌𝑔 + 𝑉1 2 2𝑔 + 𝑍1 = 𝑃2 𝜌𝑔 + 𝑉2 2 2𝑔 + 𝑍2 Carefully designed to minimize the energy losses Converging section: 𝒅𝒑 𝒅𝒙 < 𝟎 Favorable pressure gradient, Stable flow, negligible energy loss Diverging section: 𝒅𝒑 𝒅𝒙 > 0 Adverse pressure gradient: Unstable flow, energy loss Converging section (1) (2) Δh h2=P2/ρg h1=P1/ρg 𝑃1 𝜌𝑔 + 𝑧1 − 𝑃2 𝜌𝑔 + 𝑧2 = 𝑉2 2 2𝑔 − 𝑉1 2 2𝑔 ∆ℎ = 𝑄2 2𝑔 1 𝐴2 2 − 1 𝐴1 2 𝑄 = 𝐴1 𝐴2 2𝑔∆ℎ 𝐴1 2 − 𝐴2 2 Continuity equation: 𝑄 = 𝐴1 𝑉1 = 𝐴2 𝑉2; 𝐴2 < 𝐴1; 𝑉2 > 𝑉1 ; 𝑉1 = 𝐴2 2𝑔∆ℎ 𝐴1 2 − 𝐴2 2 Constriction flow meters Orifice meter Flow nozzle Venturi meter Constriction flow meters In terms of the manometer readings 𝑃1 + 𝜌𝑔𝑍1 = 𝑃2 + 𝜌𝑔(𝑍2 − ℎ) + 𝜌 𝑚𝑎𝑛 𝑔ℎ 𝑃1 − 𝑃2 𝜌𝑔 + 𝑍1 − 𝑍2 = ℎ 𝜌 𝑚𝑎𝑛 𝜌 − 1 𝑉1 = 𝐴2 2𝑔 𝑃1 − 𝑃2 𝜌𝑔 + 𝑍1 − 𝑍2 𝐴1 2 − 𝐴2 2 𝑉1 = 𝐴2 2𝑔ℎ 𝜌 𝑚𝑎𝑛 𝜌 − 1 𝐴1 2 − 𝐴2 2 𝑄 𝑎𝑐𝑡 = 𝐶 𝑑 𝐴1 𝐴2 2𝑔ℎ 𝜌 𝑚𝑎𝑛 𝜌 − 1 𝐴1 2 − 𝐴2 2 Venturi meter About 200 About 60 Z2 Z1 h Datum Streamline pattern Orifice meter RoundedSharp-edge Square shoulder Thick-plate square edge Orifice in a pipe
- 4. 4 ‘small orifice’: orifice diameter is small compared to head producing flow (head does not vary across the orifice) h P0 (1) (2) 0 open to atm 0 Tank is large 0 Torricelli’s theorem: Velocity of a issuing jet is proportional to the square root of the head producing flow Liquid flow Small orifice Applying Bernoulli’s equation to (1) and 2) 𝑃1 𝜌𝑔 + 𝑍1 + 𝑉1 2 2𝑔 = 𝑃2 𝜌𝑔 + 𝑍2 + 𝑉2 2 2𝑔 𝑉2 = 2𝑔ℎ ; 𝑄 = 𝑎 2𝑔ℎ Vena Contractor Actual area Vena Contractor Actual area In practice, actual discharge is less than the theoretical discharge Reasons: Actual velocity <Theoretical velocity due to energy loss between (1) and (2) 𝑉𝑎𝑐𝑡 = 𝐶 𝑣 2𝑔ℎ 𝐶𝑣− coefficient of velocity Fluid path converges on the orifice. Area is less than the orifice area 𝐴 𝑎𝑐𝑡 = 𝐶𝑐 𝐴 𝑜𝑟𝑖𝑓𝑖𝑐𝑒; 𝐶𝑐- coefficient of contraction 𝑄 𝑎𝑐𝑡 = 𝐶𝑣 2𝑔ℎ . 𝐶𝑐 𝐴 𝑜𝑟𝑖𝑓𝑖𝑐𝑒 𝑄 𝑎𝑐𝑡 = 𝐶𝑣 2𝑔ℎ . 𝐶𝑐 𝐴 𝑜𝑟𝑖𝑓𝑖𝑐𝑒 Coefficient of discharge 𝐶 𝑑 = 𝐶𝑐 𝐶 𝑣 Coefficient of discharge depends on the edge condition 𝐶𝑐 ≈ 1 𝐶𝑣 ≈ 0.98 𝐶 𝑑 ≈ 0.98 𝐶𝑐 ≈ 1 𝐶𝑣 ≈ 0.86 𝐶 𝑑 ≈ 0.86 𝐶𝑐 ≈ 0.62 𝐶𝑣 ≈ 0.98 𝐶 𝑑 ≈ 0.61 𝐶𝑐 ≈ 0.62 𝐶𝑣 ≈ 0.98 𝐶 𝑑 ≈ 0.61 *Contraction of jet RoundedSharp-edge Square shoulder Thick-plate square edge Determination of the coefficient of contraction, velocity, and discharge (Cc , Cv , and Cd) x y H 𝑢 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 = 2𝑔𝐻 𝑈𝐴 = 𝐶𝑣 𝑉𝑇 𝐶𝑣 = 𝑔 2𝑦 𝑥 2𝑔𝐻 𝐶𝑣 = 𝑥2 4𝑦𝐻 𝐶 𝑑 = 𝑄 𝐴 𝑄 𝑇 𝑄 𝑇 = a 2𝑔ℎ 𝑄 𝐴 √ Use direct method 𝐶𝑐 = 𝐶 𝑑 𝐶𝑣 𝑥 = 𝑢𝑡; 𝑡 = 𝑥 𝑢 𝑦 = 1 2 𝑔𝑡2 𝑢 = 𝑔 2𝑦 𝑥
- 5. 5 Falling head method: Determine coefficient of discharge (For any outflow device) 𝑄 = 𝐴𝑉 𝑄 = 𝐴 𝑑ℎ 𝑑𝑡 A 𝑑ℎ 𝑑𝑡 = 𝑄𝑖𝑛 − 𝑄 𝑜𝑢𝑡 𝑄𝑖𝑛=0; 𝑄 𝑜𝑢𝑡 = 𝐶 𝑑 𝑎 2𝑔𝐻 for small orifice A 𝑑ℎ 𝑑𝑡 = −𝐶 𝑑 𝑎 2𝑔𝐻 𝑑𝑡 = − 𝐴 𝐶 𝑑 𝑎 2𝑔 𝑑 𝑑ℎ ℎ0.5 𝑡 = − 2𝐴 𝐶 𝑑 𝑎 2𝑔 𝑑 ℎ0.5 + 𝐶 𝑡 = 0;ℎ = ℎ1 𝑡 = 𝑇; ℎ = ℎ2 𝑇 = 2𝐴 𝐶 𝑑 𝑎 2𝑔 ℎ1 0.5 − ℎ2 0.5 h1 h2 Area A Orifice area =a Exercise A large tank contains a liquid to a depth z. A small orifice located at height y above the tank base discharges a horizontal jet to the atmosphere. The jet strikes the base level of the tank at a horizontal distance x. Assume 𝐶𝑣 = 1 , 𝐶𝑐 = 0.63, 𝐴 = 670𝑎 (i) Show that 𝑥2 + 4𝑦2 = 4𝑦𝑧 and that the maximum horizontal distance occurs when 𝑧 = 2𝑦 (ii) Given y=0.25 m, find the time taken for the jet striking distance x to change from 𝑥1 = 1𝑚 to 𝑥2 = 0.5𝑚 h1 h2 x y z Exercise: Approach velocity (𝑽 𝟎) correction H=2.5 m P0 (0) (1) Liquid flow Area: A Area: a 𝑃0 𝜌𝑔 + 𝑍0 + 𝑉0 2 2𝑔 = 𝑃1 𝜌𝑔 + 𝑍1 + 𝑉1 2 2𝑔 𝑍0 + 𝑉0 2 2𝑔 =𝑍1 + 𝑉1 2 2𝑔 𝑉0 2 2𝑔 + 𝐻= 𝑉1 2 2𝑔 𝐴0 𝑉0 = 𝐴1 𝑉1 Applying continuity equation: 𝑉0 = 𝑎 𝐴 𝑉1 𝑎 𝐴 2 𝑉1 2 2𝑔 + 𝐻= 𝑉1 2 2𝑔 𝑉 = 2𝑔𝐻 1 − 𝑎2 𝐴2 A/a V 5 ? 10 ? 100 ? 1000 ? 𝑉 = 2𝑔𝐻 = 7 𝑚/𝑠 Assumed small area; if orifice area is not small h1 h2 𝑉1 = 2𝑔ℎ1 𝑉2 = 2𝑔ℎ2 Velocity will change across the orifice Head vary substantially from top to bottom
- 6. 6 For small area dA; 𝑉 = 2𝑔ℎ 𝑑𝑄 = 𝑏𝑑ℎ 2𝑔ℎ Large rectangular orifice: vertical height is large h1 h2 h dh *Side contraction Integrate over area; 𝑄 = 𝑏 2𝑔 ℎ. 𝑑ℎ ℎ2 ℎ1 𝑄 = 𝐶 𝑑 2𝑏 3 2𝑔 ℎ2 3/2 − ℎ1 3/2 Notches and weirs Notch: Opening in the side of a tank or reservoir, extending above the free surface (No upper edge) Weir: notch on a large scale, to measure the flow of a river Rectangular notch b H h 𝑄 = 𝐶 𝑑 2𝑏 3 2𝑔 𝐻3/2 𝑄 = 𝐶 𝑑 2𝑏 3 2𝑔 ℎ2 3/2 − ℎ1 3/2 0 𝑑𝐴 = 2 𝐻 − ℎ 𝑇𝑎𝑛𝜃. 𝑑ℎ 𝑑𝑄 = 2𝑔ℎ .2 𝐻 − ℎ 𝑇𝑎𝑛𝜃. 𝑑ℎ 𝑑𝑄 = 2 2𝑔 𝑡𝑎𝑛𝜃. ℎ1/2 𝐻 − ℎ 𝐻 0 𝑑ℎ 𝑄 𝑎𝑐𝑡 = 𝐶 𝑑 8 15 2𝑔 𝑡𝑎𝑛𝜃 𝐻5/2 Vee notch H h Weir Calibration For a V notch: Applying mass continuity equation (assume quasi steady condition) 𝐴 𝑑𝐻 𝑑𝑡 = 𝑄𝑖𝑛 − 𝑄 𝑜𝑢𝑡 = 0 − 𝐶 𝑑 8 15 2𝑔 𝑡𝑎𝑛𝜃 𝐻 5 2 Integrating: 𝑇 = 5𝐴 4𝐶 𝑑 𝐻2 −3/2 − 𝐻1 −3/2 𝑡𝑎𝑛𝜃 2𝑔
- 7. 7 Angular momentum equation Moment of momentum equation For a particle dm having velocity 𝑉 ; Linear momentum = 𝑉.dm Moment about axis on O = 𝑟 x 𝑉.dm 𝐻0 = 𝑟 x 𝑉.dm 𝑠𝑦𝑠 H- moment of momentum 𝑉𝑡 𝑇𝑟𝑒𝑛𝑠𝑣𝑒𝑟𝑠𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑉𝑟 𝑅𝑎𝑑𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 Moment of momentum Angular momentum equation Newton’s 2nd law to rotating systems: 𝐹 = 𝑑 𝑑𝑡 𝑚𝑉 𝑟x 𝐹 = 𝑟 𝑑 𝑑𝑡 𝑚𝑉 𝑟x 𝐹 = 𝑑 𝑑𝑡 𝑟 × 𝑚 𝑉 𝑇 = 𝑑𝐻 𝑑𝑡 T net torque or moment applied on the system Rate of change of moment of momentum (angular momentum) of a system is equal to the net torque acting on the system Angular momentum equation-CV Expression 𝐻0 = 𝑟 x 𝑉.dm 𝑠𝑦𝑠 𝑇 = 𝑑𝐻 𝑑𝑡 Newton’s 2nd law to rotating systems 𝑑𝐵𝑠𝑦𝑠 𝑑𝑡 = 𝜕 𝜕𝑡 𝛽𝜌𝑑∀ 𝐶𝑉 + 𝛽 𝐶𝑆 𝜌𝑉. 𝑑𝐴 For a stationary CV, RTT 𝑑𝐵 = 𝛽. 𝑑𝑚 = 𝛽. 𝜌. 𝑑∀ 𝐵 = 𝛽. 𝜌. 𝑑∀ 𝐶𝑉 Extensive property B = moment of momentum 𝐵 = 𝑟 x 𝑉.dm = 𝑟 x 𝑉.ρ𝑑∀𝐶𝑉𝐶𝑉 𝛽 = 𝑑𝐵 𝑑𝑚 = 𝑑𝐻0 𝑑𝑚 = 𝑟 x 𝑉 CV expression for angular momentum equation 𝑑𝐻 𝑠𝑦𝑠 𝑑𝑡 = 𝜕 𝜕𝑡 𝑟x 𝑉𝜌𝑑∀𝐶𝑉 + 𝑟 × 𝑉𝐶𝑆 𝜌𝑉. 𝑑𝐴 𝑇 = 𝑑𝐻 𝑑𝑡 𝐶𝑉 + 𝐻 𝑜 𝑜𝑢𝑡 -𝐻 𝑜 𝑖𝑛 Angular momentum equation-CV Expression Cont. Steady flow 𝑑𝐻 𝑑𝑡 𝐶𝑉 =0 ; 𝑇 = 𝐻 𝑜 𝑜𝑢𝑡 −𝐻 𝑜 𝑖𝑛 = (𝑟 × 𝑉𝑚) 𝑜𝑢𝑡- (𝑟 × 𝑉𝑚)𝑖𝑛 Steady 1D flow: 𝑇 = ρ𝑄 𝑟2 𝑉2 − 𝑟1 𝑉1 The sum of all external moments acting on a CV The rate of change of the angular momentum of the contents of the CV The net rate of angular momentum out of the control surface = + 𝑇 = 𝑑𝐻 𝑑𝑡 𝐶𝑉 + 𝐻 𝑜 𝑜𝑢𝑡 -𝐻 𝑜 𝑖𝑛
- 8. 8 Application Pump 𝑇 = ρ𝑄 𝑟2 𝑉2 − 𝑟1 𝑉1 Fluid EnergyMechanical Energy Centrifugal Pump Flow direction Vt2 Vt1 r2 r1 ω Example 1 20 cm 30 cm A Bω A sprinkler with two nozzles of diameter 4 mm each is connected across a tap of water. The nozzles are at a distance of 30 cm and 20 cm from the centre of the tap. The flow rate of water through tap is 120 cm3/s. The nozzles discharge water in the downward direction. Determine the angular speed at which the sprinkler will rotate free. 2- arm rotating sprinkler 𝜃 Datum z=0 Total head Piezometric head line 𝑃1 𝜌𝑔 𝑉1 2 2𝑔 𝑍1 𝑍2 𝑃2 𝜌𝑔 𝑉2 2 2𝑔 𝐴1 𝐴2 𝑃 𝜌𝑔 + 𝑧 + 𝑉2 2𝑔 = 𝐻 𝑃1 𝜌𝑔 + 𝑍1 + 𝑉1 2 2𝑔 = 𝑃2 𝜌𝑔 + 𝑍2 + 𝑉2 2 2𝑔 Consider a streamline between two points 1 and 2 Bernoulli equation Modifications of Bernoulli Equation In practice, the total energy of a streamline does not remain constant Consider a streamline between two points 1 and 2 𝐻1 + 𝐻𝑒𝑎𝑑 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛𝑠 − 𝐻𝑒𝑎𝑑 𝑙𝑜𝑠𝑠𝑒𝑠 = 𝐻2 Head balance equation 𝐻1 ∓ 𝐻 𝐸 − 𝐻𝑓 = 𝐻2 𝑃1 𝜌𝑔 + 𝑉1 2 2𝑔 + 𝑍1 ∓ 𝐻 𝐸 − 𝐻𝑓 = 𝑃2 𝜌𝑔 + 𝑉2 2 2𝑔 + 𝑍2 (3.12) Energy is ‘lost’ through friction (actually transformed into heat), and external energy may be either added by means of a pump or extracted by a turbine 𝜃𝐴1 𝐴2 Losses (transformed into another state) Additions
- 9. 9 ℎ 𝑓 = 𝜆 𝐿 𝐷 𝑉2 2𝑔 L (m) to to (N/m2) to to v (m/s) P1 (N/m2) P2 (N/m2) 1 2 D (m) Head loss due to friction If the flow is inviscid ℎ 𝑓 = 0 The effect of friction on layers of water molecules Viscosity: measure of a fluid's resistance to flow Sudden Expansion Sudden Contraction The flow is not able to follow the shape. As a result, there is flow separation, creating turbulent eddies. Minor loss: energy loss due to obstructions hL=K . (v2 2/2g) hL=K . (v1 2/2g) hL=KL . (v2 2/2g)Entrance losses: Minor loss: energy loss due to obstructions hL=KL . (v1 2/2g)Exit losses: Minor loss: energy loss due to obstructions
- 10. 10 Minor loss: energy loss due to obstructions A divergent duct or diffuser Tee junctions Bends Y junctions P = power added to fluid 𝑃 = 𝜌𝑔 𝑄 𝐻 𝐸 Pumps add energy to the flow Mechanical Energy Fluid Energy Centrifugal Pump Turbines extracts energy from the flow Mechanical Energy Fluid Energy Pelton Wheel Impeller Buckets Inlet Nozzle Spear Discharge Deflector plate 2 2 V g 2 2 V g p g Representation of energy terms 2 2 V g p g wh wh wh 2 2V g p g 2 2 V g p g Representation of energy terms
- 11. 11 Conservation of Energy 𝑑𝐸𝑆 𝑑𝑡 = 𝐻𝑡 + 𝑊𝑘 Rate of change of energy of a system Net rate of heat transfer Net power (rate of work) input to the system Energy (E) = kinetic energy+ potential energy+ internal energy (related to molecularmotion) = 1 2 𝑚𝑉2 + 𝑚𝑔𝑍 + 𝑚𝑖 Specific energy (e )= 1 2 𝑉2 + 𝑔𝑍 + 𝑖 Incompressible fluids with constant temperature: 𝑑𝐸 𝑆 𝑑𝑡 = 𝑊𝑘 E = 1 2 𝑚𝑉2 + 𝑚𝑔𝑍; e = 1 2 𝑉2 + 𝑔𝑍 In the analysis of fluid flow systems: 𝑬 𝟏 + 𝑬𝒏𝒆𝒓𝒈𝒚 𝒂𝒅𝒅𝒊𝒕𝒊𝒐𝒏𝒔 − 𝑬𝒏𝒆𝒓𝒈𝒚 𝒍𝒐𝒔𝒔𝒆𝒔 = 𝑬 𝟐 𝑬 𝟏 − 𝑬 𝟐 = 𝑬𝒏𝒆𝒓𝒈𝒚 𝒂𝒅𝒅𝒊𝒕𝒊𝒐𝒏𝒔 − 𝑬𝒏𝒆𝒓𝒈𝒚 𝒍𝒐𝒔𝒔𝒆𝒔 𝒅𝑬 = 𝑬𝒏𝒆𝒓𝒈𝒚 𝒂𝒅𝒅𝒊𝒕𝒊𝒐𝒏𝒔 − 𝑬𝒏𝒆𝒓𝒈𝒚 𝒍𝒐𝒔𝒔𝒆𝒔 Internal energy: Energy due to molecular motion in a fluid, depends on the temperature and physical state (liquid, gas, two phase) 𝜃𝐴1 𝐴2 𝐸1 𝐻1 𝐸2 𝐻2 Losses Additions Work associated with force acting through a distance 𝑊𝑇𝑜𝑡𝑎𝑙 = 𝑊𝑠ℎ𝑎𝑓𝑡 + 𝑊𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 + 𝑊𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑊𝑠ℎ𝑎𝑓𝑡- work transmitted by a rotating shaft (Turbine, pumps) 𝑊𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒-work done by the pressure forces on the control surface 𝑊𝑣𝑖𝑠𝑐𝑜𝑢𝑠- normal and shear components of the viscous forces on the control surface Note: not considered if the moving walls are inside the control volume and fixed walls are outside the control volume 𝑊𝑇𝑜𝑡𝑎𝑙 = 𝑊𝑠ℎ𝑎𝑓𝑡 + 𝑊𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 Power transmitted through rotating shaft 𝑊𝑠ℎ𝑎𝑓𝑡 = 𝜌𝑔𝑄𝐻 = 𝜔𝑇𝑠ℎ𝑎𝑓𝑡 = 2𝜋𝑛 𝑇𝑠ℎ𝑎𝑓𝑡 𝜔 –angular speed of shaft (rad/s) 𝑇𝑠ℎ𝑎𝑓𝑡-Shaft torque 𝑛 - number of revolutions of the shaft per unit time (rpm) Work done by the pressure=Force x Distance travelled = 𝑃𝐴 × 𝑚 𝜌 𝐴 = 𝑃 × 𝑚 𝜌 Incompressible fluids with constant temperature: 𝑑𝐸𝑆 𝑑𝑡 = 𝑊𝑘 = 𝑊𝑠ℎ𝑎𝑓𝑡 + 𝑊𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 E = 1 2 𝑚𝑉2 + 𝑚𝑔𝑍; e = 1 2 𝑉2 + 𝑔𝑍 𝑑𝐵𝑠𝑦𝑠 𝑑𝑡 = 𝜕𝐵𝑐 𝜕𝑡 + 𝐵 𝑜𝑢𝑡 − 𝐵𝑖𝑛 = 𝜕 𝜕𝑡 𝛽𝜌𝑑∀ 𝐶𝑉 + 𝛽 𝐶𝑆 𝜌𝑉. 𝑑𝐴 B: Energy (E) 𝑑𝐸𝑠𝑦𝑠 𝑑𝑡 = 𝜕𝐸𝑐 𝜕𝑡 + 𝐸 𝑜𝑢𝑡 − 𝐸𝑖𝑛 = 𝜕 𝜕𝑡 𝑒𝜌𝑑∀ 𝐶𝑉 + 𝑒 𝐶𝑆 𝜌𝑉. 𝑑𝐴 β: Energy (E)/unit mass = e RTT; 𝑆𝑡𝑒𝑎𝑑𝑦 1𝐷 𝑓𝑙𝑜𝑤 ; 𝜕𝐸 𝑐 𝜕𝑡 = 0 𝑑𝐸𝑠𝑦𝑠 𝑑𝑡 = 𝐸 𝑜𝑢𝑡 − 𝐸𝑖𝑛 = 𝑒 𝐶𝑆 𝜌𝑉. 𝑑𝐴 𝑑𝐸𝑠𝑦𝑠 𝑑𝑡 = ( 1 2 𝑚 𝑉2 + 𝑚 𝑔𝑍 + 𝑃 𝑚 𝜌 ) 𝑜𝑢𝑡−( 1 2 𝑚 𝑉2 + 𝑚 𝑔𝑍 + 𝑃 𝑚 𝜌 )𝑖𝑛= ( 1 2 𝑉2 + 𝑔𝑍 + 𝑃 𝑚 𝜌 ) 𝐶𝑆 𝜌𝑉. 𝑑𝐴 𝑊𝑠ℎ𝑎𝑓𝑡/𝑚 𝑔 = ( 𝑉2 2𝑔 + 𝑍 + 𝑃 𝜌𝑔 ) 𝑜𝑢𝑡−( 𝑉2 2𝑔 + 𝑍 + 𝑃 𝜌𝑔 )𝑖𝑛 𝑊𝑠ℎ𝑎𝑓𝑡 𝑚 𝑔 = 2𝜋𝑛 𝑇𝑠ℎ𝑎𝑓𝑡 𝑚 𝑔 = 𝜌𝑔𝑄𝐻 𝑚 𝑔 = 𝐻 𝑊𝑠ℎ𝑎𝑓𝑡/𝑚 𝑔 = ( 𝑉2 2𝑔 + 𝑍 + 𝑃 𝜌𝑔 ) 𝑜𝑢𝑡−( 𝑉2 2𝑔 + 𝑍 + 𝑃 𝜌𝑔 )𝑖𝑛 𝐻 𝐸 = ( 𝑉2 2𝑔 + 𝑍 + 𝑃 𝜌𝑔 ) 𝑜𝑢𝑡−( 𝑉2 2𝑔 + 𝑍 + 𝑃 𝜌𝑔 )𝑖𝑛
- 12. 12 A large closed tank contains a liquid of density 850 kg/m3 and air is under pressure of 25 kPa. The liquid is discharged to the atmosphere at N through nozzle of diameter 60 mm located at the end of pipe of diameter 120 mm fitted with a pump as shown in the Figure. The head loss in the pipe AP (suction pipe) is 6 𝑉2 2𝑔 and the head loss in the pipe PM (delivery pipe) is 4 𝑉2 2𝑔 where V is the velocity in pipes JP and PM. The head loss in nozzle MN is negligible. Find the power required by the pump to deliver 25 l/s. Draw the total head line clearly indicating magnitudes of changes in head. Find the discharge, if the head added by the pump drops by 50 %. M Water is pumped from reservoir A to reservoir B through a pipe with the highest point C at a flow of 60 l/s. The diameter of the pipe is 0.2 m. The atmospheric pressure is 10 m water. The head losses hL between C and B are 2.8 m. To what highest elevation at point C can the pipe reach if the absolute pressure has to be at least 2 m of water. pump A B C +15 +30 20 m 300 m 100 m A large open tank contains an oil of density 850 kg/m3. The oil is discharged to the atmosphere at K through a pipeline fitted with a pump at P. The head loss in the 150 mm diameter pipe JP is 2 𝑉1 2 2𝑔 and the head loss in the 100 mm diameter pipe PK is 3 𝑉2 2 2𝑔 where V1 and V2 are the velocities in pipes JP and PK, respectively. a) When the pump adds 5 kW of power causing a discharge of 60 l/s at K, find the elevation h at K and the deflection δ of the mercury manometer What is the volume flow rate through the pipe if the pump is removed? A large closed tank contains a liquid of density 850 kg/m3 and air is under pressure P0. The liquid is discharged to the atmosphere through a pipeline fitted with a pump. The head loss in the 300 mm diameter pipe JM is 9 𝑉2 2𝑔 .The head loss in pipe MN is negligible. For Q= 140 l/s, find P0 when the pump is not working.
- 13. 13 A large open tank contains an oil of density 850 kg/m3. The tank is discharged to the atmosphere at K through a pipeline fitted with a pump at P as shown in figure. The head loss in the 150 mm diameter pipe JP is 2 𝑉1 2 2𝑔 and the head loss in the 100 mm diameter pipe PK is 3 𝑉2 2 2𝑔 where V1 and V2 are the velocities in pipes JP and PK, respectively.